### (0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

mul0(Cons(x, xs), y) → add0(mul0(xs, y), y)
add0(Cons(x, xs), y) → add0(xs, Cons(S, y))
mul0(Nil, y) → Nil
add0(Nil, y) → y
goal(xs, ys) → mul0(xs, ys)

Rewrite Strategy: INNERMOST

### (1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0
goal(z0, z1) → mul0(z0, z1)
Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
MUL0(Nil, z0) → c1
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
ADD0(Nil, z0) → c3
GOAL(z0, z1) → c4(MUL0(z0, z1))
S tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
MUL0(Nil, z0) → c1
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
ADD0(Nil, z0) → c3
GOAL(z0, z1) → c4(MUL0(z0, z1))
K tuples:none
Defined Rule Symbols:

Defined Pair Symbols:

Compound Symbols:

c, c1, c2, c3, c4

### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

GOAL(z0, z1) → c4(MUL0(z0, z1))
Removed 2 trailing nodes:

ADD0(Nil, z0) → c3
MUL0(Nil, z0) → c1

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0
goal(z0, z1) → mul0(z0, z1)
Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
S tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
K tuples:none
Defined Rule Symbols:

Defined Pair Symbols:

Compound Symbols:

c, c2

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

goal(z0, z1) → mul0(z0, z1)

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0
Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
S tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
K tuples:none
Defined Rule Symbols:

Defined Pair Symbols:

Compound Symbols:

c, c2

### (7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ADD0(x1, x2)) = [4]
POL(Cons(x1, x2)) = [5] + x2
POL(MUL0(x1, x2)) = [2]x1
POL(Nil) = [5]
POL(S) = 0
POL(add0(x1, x2)) = [4] + [2]x2
POL(c(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(mul0(x1, x2)) = [2] + [3]x2

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0
Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
S tuples:

ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
K tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
Defined Rule Symbols:

Defined Pair Symbols:

Compound Symbols:

c, c2

### (9) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^3))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
We considered the (Usable) Rules:

add0(Nil, z0) → z0
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
mul0(Nil, z0) → Nil
mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
And the Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ADD0(x1, x2)) = [1] + x1
POL(Cons(x1, x2)) = [1] + x2
POL(MUL0(x1, x2)) = x13 + x12·x2
POL(Nil) = 0
POL(S) = 0
POL(add0(x1, x2)) = x1 + x2
POL(c(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(mul0(x1, x2)) = x1·x2

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

mul0(Cons(z0, z1), z2) → add0(mul0(z1, z2), z2)
mul0(Nil, z0) → Nil
add0(Cons(z0, z1), z2) → add0(z1, Cons(S, z2))
add0(Nil, z0) → z0
Tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
S tuples:none
K tuples:

MUL0(Cons(z0, z1), z2) → c(ADD0(mul0(z1, z2), z2), MUL0(z1, z2))
ADD0(Cons(z0, z1), z2) → c2(ADD0(z1, Cons(S, z2)))
Defined Rule Symbols:

Defined Pair Symbols: